Electrical igniter for gas engines A keystone to understanding by Magluvin

[Montec]

Hello Magluvin

Since you are basically moving energy from one capacitor to another it seems that you need to think in units of energy and not in units of volts.

Energy in a capacitor is U = 1/2(CV²)

   V = applied volts
   C = capacitance

BTW the 1/2 comes out of the integral used to calculate the area under the capacitor charge curve vs time period.

Hyperphysics has a good explanation http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html


 

[woopy]

Hi all

the real thing is if we apply the  formula, which is written in all text book, the energy transfer from a cap to a cap of same value is 50 % efficient.

and of course the question is where did those other 50% go

In heat or something else ? in this case the 50 % energy  are fully lost and unrecoverable yes or not . and in this case it is simply not possible to recover them or a part of them , because they are no more there ?? yes or not ?

So my question is , from where  the energy of the Inductor and diode ( "believe circuit"  as Mag named it ) comes from , to recreate much better efficiency in THE SAME ENERGY TRANSFER.

My bench experiment shows around 50 % efficiency in a direct transfer, that is transferring a cap to a cap without any resistance       and about  85 % efficiency by ADDING RESISTANCE OF INDUCTOR AND DIODE

Please explanation

Hi wayne
i do not understand your 200 % gain in your calculation, for me the result is the same , if i charge 2 cap of 10 uF at 10 volts, they store the same energy as those same charged cap mounted in serie     of course as per the bloody 1/2 X CAPACITY X VOLTAGE ^2.

good luck at all

Laurent

[woopy]

Hi Montec


thank's for the post concerning the , as i named it " bloody formula " there is no insult here against the formulator people but  this formula puzzle me and really derange me on this research.

I know that this formula is used all over the place concering the kinetic energy and it is in general used with atmost efficacity .

as i can understand ,  this formula says that a single cap of  10 uF and 10 volts , is discharged in a same empty cap of 10 uF .and by doing this it loses 50 % of the stored energy.(as per hyperfhisics link )

So i can understand that the voltage is a pressure or a spring. So at 10 volts the spring is streched with a lot of force, or work. Than we decide to split in the same capacity the stored force, so the concentrated  energy will be diluated in 2 entities.  So you you simply distreched the spring to release  more space, in order to refill the 2 capacities at half the voltage. or half the spring tension .

So more volume at less concentration.  and of course because the bloody formula  50 % lost energy.

OK but now we are at the end of the  releasing phase and every thing is calm and equalised... but where are the bloody energy come from and says   " please wake up everybody , we can not stay at this point  and we will pile  all above the other and recreate some more energy. So we will rebuild from 50%  of the standard transfer up to 85 % or more.

where is this staff of people comes from ?

sorry to repeat me , but explanation and discussion is welcome

good luck at all

laurent

[poynt99]

Laurent,

I assume you are in general, "OK" with the equation in question, because you have seen first-hand on your bench that the equation seems to be true to your results. That is, when you use no inductor, and have a pure resistance between the two capacitors, energy seems to be "lost".

You are having difficulty accepting it, but it is true, when using a piece of wire, 50% of the energy will be lost in heat in that interconnecting wire.

When you insert an inductance (and a diode) between the two capacitors, the situation is now different. Now, rather than a dissipative element (the piece of resistive wire), you have a less dissipative element (an inductance and lump resistance) between the two capacitors. This less dissipative element stores some of the energy that otherwise would be lost if only a wire was used. The inductor not only stores some of this energy, but it releases it again in the direction of the empty capacitor.

You can continue to improve the efficiency of this energy transfer by increasing the inductance, while trying to maintain a relatively low lumped resistance.

Eventually, you will approach 100% energy transfer, but you will not quite get there, due to the finite resistance we always have in inductors (not super-cooled).

Hope that helps.

.99

[woopy]

 Thank's Point99

i have to go sleeping

and i will be dreaming of this

in my mind,  a wire of respectable diameter  (probably 0.8 to 1mm diameter )as the one comming out of my cap, can probably transfer  some amps in a very short time, in other words a lot of energy.

the transfer of the very low power cap  (10 uF at 10 volts that is to say 500 uj)
  this is very low energy can be transported really easily through the wire without really heat creation.

So my question is     if i measure the resistance of the 1mm diameter  (Cap wire ) on the 5 centimeter length of this wire , none of my meter will be able to detect a so small  resistance,

But if i measure the resistance of the inductor (the secondary winding of a MOT) i get 107 ohms .

So adding  "a huge resistor" under the form of an inductor has a better efficiency  than a bare solid wire ( with almost no resistance) ??

I am very open to all explanations

good night at all

Laurent